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Confusion

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As a non-mathematician, but engineer, this article is somewhat confusing. It indiscriminately mixes chat about any base logarithms in with that of the natural base logs, instead of focusing on the natural logs and how they are different from the others. It's missing what I recall as the defining characteristic of the natural logs: "But the one property that goes to the essence of e and makes it so natural for logarithms and situations of exponential growth and decay is this: d(e^x/dx) = e^x." 74.127.200.33 (talk) 18:45, 11 October 2022 (UTC)[reply]

Nor does the article explain why such logarithms are needed or what advantages they offer. Carusus (talk) 12:30, 5 January 2025 (UTC)[reply]

the discussion on units needs to be returned

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It is false to say that the argument of a logarithm must be unitless. There is no mathematics to support this assertion. Engineers, physicist and chemists routinely take the logarithm of quantities with units and that most certainly does not violate any mathematical concepts. The comment:

this is false, that integral only works because it's a definite integral evaluating to log(b/a), i.e. we still cancelled units to make the inside unitless

Is irrelevant and false. ALL NATURAL LOGARITHMS ARE DEFINITE INTEGRALS. There is no such thing as a natural log that is not a definite integral. The divisor of a (to remove the units) in the post text above is incorrect and unneeded. The mathematical PROOF I provided shows that units are allowed in the argument of log function and that the resulting value is always unitless. Unless an editor can provide a mathematical PROOF that it is not true then the section on units should be returned.

here is the removed text:

Units

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An often asked question is "what happens if the argument to the log function (natural or any other base) has units"? A Google search will likely provide the incorrect answer and return many pages that say the argument to a log function must be unitless. This is incorrect. The argument can, and often does, have units. However, the value returned by a log function is always unitless regardless of the units of the argument.


The simplest way to understand how the log function handles units is by examining the definition for the natural log as an Integral. The natural log of is the area under the curve of plotted against in the region from 1 to . Note that the units of an integral are always the units of the x-axis times the units of the y-axis. But since for the logarithm the y-axis is defined as , the resulting units of the area are and the units always cancel and the area is unitless. So the natural logarithm of a quantity with units is always unitless regardless of the unit of the quantity. Furthermore, there are no restrictions on the units of the argument, it can be unitless or have any unit desired.


It does not matter which mathematical definition is used for the natural log since all definitions are mathematically equivalent. Furthermore, in mathematics, units are treated exactly the same way the numeric values are. A mathematical equation must work with units or the mathematics is wrong.


A corollary of the way a logarithm handles units is that the log function is a "lossy function" and the inverse function can not return the original quantity. This is similar to the way squaring works. The inverse function, square root, can not return the original value of the function since squaring a number removes the sign, and the square root can not recover that lost sign. So the for all . For a logarithm, the exponent; for all since the log function has removed the units of and exponentiation can not recover the lost units. For example, .

Calculator

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Regarding recent edits adding and then removing a calculator gadget from this article: See discussion at Talk:Binary_logarithm#Calculator, which I started there before realizing that this article was also involved. —David Eppstein (talk) 07:21, 20 January 2025 (UTC)[reply]

Plot caption

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The plot under the function plot reads, in part, “... slowly goes to negative infinity as x approaches 0.” This is clearly a misstatement, as the function goes to negative infinity very rapidly. Marktomory (talk) 20:17, 17 March 2025 (UTC)[reply]

As precised just after that:" "slowly" as compared to any power law of x". --Sapphorain (talk) 20:32, 17 March 2025 (UTC)[reply]